Question

A chemistry student weighs out 0.0475g of sulfurous acid H2SO3, a diprotic acid, into a

Answer 1

I don't know but look on the internet or use a calculator 

Answer 2

Answer: The volume of NaOH required is 10.5 mL

Explanation:

• For sulfurous acid:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Mass of solute (sulfurous acid) = 0.0475 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250. mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.0475g\times 1000}{82g/mol\times 250.mL}\\\\\text{Molarity of sulfurous acid}=2.32\times 10^{-3}M$

• To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=2.32\times 10^{-3}M\\V_1=250.mL\\n_2=1\\M_2=0.1100M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 2.32\times 10^{-3}\times 250.=1\times 0.1100\times V_2\\\\V_2=10.5mL$

Hence, the volume of NaOH required is 10.5 mL